The sum of the first 15 terms of the A.P. with \(a = -3\) and \(d = 4\) is
A. \(405\)
B. \(345\)
C. \(750\)
D. \(375\)
A. \(405\)
B. \(345\)
C. \(750\)
D. \(375\)
Best answer
Answer: D. \(375\)
\(S_{15} = \frac{15}{2}[2(-3) + 14 \times 4] = \frac{15}{2}(-6 + 56) = \frac{15}{2} \times 50 = 375\).
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